TITLE: I'm trying to calculate $e^{At}$. Where do I go wrong?
QUESTION [7 upvotes]: Let $$A=\begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -4 & 0\end{bmatrix}$$
I want to determine $e^{At}$. I tried it using two methods.
First via the Jordan form:
The characteristic polynomial of $A$ is given by $$\det(I\xi-A)=\begin{vmatrix}
\xi & -1 & 0 & 0 \\ 1 & \xi & 0 & 0 \\ 0 & 0 & \xi & -1 \\ 0 & 0 & 4 & \xi
\end{vmatrix}=1\cdot\begin{vmatrix}
1 & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi
\end{vmatrix}+\xi\cdot\begin{vmatrix}
\xi & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi
\end{vmatrix}$$
$$\begin{vmatrix}
1 & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi
\end{vmatrix}=1\cdot\begin{vmatrix}
\xi & -1 \\ 4 & \xi
\end{vmatrix}=\xi^{2}+4\qquad\text{and}\qquad \begin{vmatrix}
\xi & 0 & 0 \\ 0 & \xi & -1 \\ 0 & 4 & \xi
\end{vmatrix}=\xi\cdot\begin{vmatrix}
\xi & -1 \\ 4 & \xi
\end{vmatrix}=\xi(\xi^{2}+4)$$
$$\implies\det(I\xi-A)=\xi^{2}+4+\xi(\xi^{3}+4\xi)=\xi^{4}+5\xi^{2}+4=(\xi^{2}+4)(\xi^{2}+1)$$ Then the eigenvalues are given by $\lambda_{1,2}=\pm i$ and $\lambda_{3,4}=\pm 2i$. Now, computing the corresponding eigenvectors:
For $\lambda=i$, we have $$\begin{cases}
-ix+y=0 \\ -x-iy=0 \\ iz+w=0 \\ -4z-iw=0
\end{cases}\implies\begin{cases}
x=-y \\ 4z=-w
\end{cases}$$ Then setting $x=a$ and $z=b$, for $a,b\in\mathbb{C}$, we obtain:
$$v_{1}=\begin{pmatrix}
a \\ -a \\ b \\ -4b
\end{pmatrix}=a\begin{pmatrix}
1 \\ -1 \\ 0 \\ 0
\end{pmatrix}+b\begin{pmatrix}
0 \\ 0 \\ 1 \\ -4
\end{pmatrix}$$
Similarly, for $\lambda=-i$, we compute, for complex $c,d$: $$v_{2}=c\begin{pmatrix}
-1 \\ 1 \\ 0 \\ 0
\end{pmatrix}+d\begin{pmatrix}
0 \\ 0 \\ -1 \\ 4
\end{pmatrix}$$
For $\lambda=2i$, we get: $$v_{3}=j\begin{pmatrix}
1 \\ -1 \\ 0 \\ 0
\end{pmatrix}+k\begin{pmatrix}
0 \\ 0 \\ 1 \\ -2
\end{pmatrix}$$
And finally, for $\lambda=-2i$: $$v_{4}=m\begin{pmatrix}
-1 \\ 1 \\ 0 \\ 0
\end{pmatrix}+n\begin{pmatrix}
0 \\ 0 \\ -1 \\ 2
\end{pmatrix}$$
Define $S=\begin{bmatrix}
v_{1} & v_{2} & v_{3} & v_{4}
\end{bmatrix}$. Then $$S=\begin{bmatrix}
a & -c & j & -m \\ -a & c & -j & m \\ b & -d & k & -n \\ -4b & 4d & -2k & 2n
\end{bmatrix}$$
But I have to calculate $S^{-1}$ via this method (which seems a bit too complicated), since $e^{At}=Se^{S^{-1}ASt}S^{-1}$.
Then I tried to compute $e^{At}$ using the theory of autonomous behaviours:
So I said that every strong solution of $\frac{d}{dt}x=Ax$ is of the form $$x(t)=B_{10}e^{-it}+B_{20}e^{it}+B_{30}e^{-2it}+B_{40}e^{2it}$$
The vectors $B_{ij}$ should satisfy the relations $$(-i\cdot I-A)B_{10}=0$$ $$(i\cdot I-A)B_{20}=0$$ $$(-2i\cdot I-A)B_{30}=0$$ $$(2i\cdot I-A)B_{40}=0$$
Solving these equations yield $B_{10}=v_{1}$, $B_{20}=v_{2}$, $B_{30}=v_{3}$, $B_{40}=v_{4}$, as calculated previously.
Hence every solution of $x$ can be written as $$x(t)=a\begin{bmatrix}-1 \\ 1 \\ 0 \\ 0\end{bmatrix}e^{-it}+b\begin{bmatrix}0 \\ 0 \\ -1 \\ 4\end{bmatrix}e^{-it}+c\begin{bmatrix}-1 \\ 1 \\ 0 \\ 0 \end{bmatrix}e^{it}+d\begin{bmatrix} 0 \\ 0 \\ 1 \\ 4 \end{bmatrix}e^{it} +j\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}e^{-2it}+k\begin{bmatrix} 0 \\ 0 \\ -1 \\ 2 \end{bmatrix}e^{-2it}+m\begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}e^{2it}+n\begin{bmatrix}0 \\ 0 \\ 1 \\ -2 \end{bmatrix}e^{2it}$$
From this we can obtain four linearly independent solutions:
$$x_{1}(t)=\begin{bmatrix}-e^{it} \\ e^{it} \\ 0 \\ 0\end{bmatrix}\qquad x_{2}(t)=\begin{bmatrix}e^{2it} \\ -e^{2it} \\ 0 \\ 0\end{bmatrix}\qquad x_{3}(t)=\begin{bmatrix}0 \\ 0 \\ e^{it} \\ 4e^{it}\end{bmatrix} \qquad x_{4}(t)=\begin{bmatrix} 0 \\ 0 \\ e^{2it} \\ -2e^{2it}\end{bmatrix}$$
The matrix $X$ is defined as $X=\begin{bmatrix}x_{1} & x_{2} & x_{3} & x_{4}\end{bmatrix}$
Then $e^{At}=\Phi(t):=X(t)X^{-1}(0)=\text{something really complicated, so I think I went wrong somewhere}$

REPLY [2 votes]: One problem for sure is how you treat complex linear systems. Look at your solution:
$$
\begin{cases}
-ix+y=0 \\ -x-iy=0 \\ iz+w=0 \\ -4z-iw=0
\end{cases}\implies\begin{cases}
x=-y \\ 4z=-w
\end{cases}
$$
If I take $x=-y=1$, for example, I will get the first equation
$$
-i-1=0.
$$
It is clearly false. The solution is $x=-iy$, $w=z=0$, so an eigenvector is, for example,
$$
\left[\matrix{-i\\1\\0\\0}\right].
$$
P.S. Note that all eigenvalues are distinct, so you cannot get more than one linear independent eigenvector for each.